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## Linear Integrated MCQ Set 1

1. Which among the following circuit has the highest input resistance?
a) Voltage follower
b) Inverting amplifier
c) Differential amplifier
d) None of the mentioned

Answer: a [Reason:] Voltage follower has the highest positive input resistance of any op-amp circuit. For this reason it is used to reduce voltage error caused by source loading.

2. Find the bias current from the given circuit a) 30mA
b) 3mA
c) 0.30mA
d) 0.03mA

Answer: c [Reason:] The bias current is given as Iin =Vin/Rin = 3v/10kΩ. Where, Iin= Ib =0.3mA.

3. How to choose an op-amp when working with high input source resistance?
a) Op-amp with low bias current
b) Op-amp with higher slew rate
c) Buffer or voltage follower
d) All of the mentioned

Answer: d [Reason:] When the op-amp is driven by a high input source resistance, the output and input voltage will not be equal due to error at the input. A remedy to this problem is an op-amp with low input bias current and high slew rate should be chosen as a voltage follower.

4. What must be done to block the ac input voltage riding on a dc level?
a) Use RC network
b) Use coupling capacitor
c) Use resistive transducer
d) None of the mentioned

Answer: b [Reason:] In order to block the dc level a coupling capacitor must be used in series with the input of the voltage follower.

5. To get higher input resistance in AC coupled voltage follower,
a) The output resistance is bootstrapped
b) The input resistance is bootstrapped
c) The bias resistance is bootstrapped
d) The feedback resistance is bootstrapped

Answer: c [Reason:] Bias resistor connected to ground to provide path in an AC coupled voltage follower, drastically reduces the input resistance of the circuit. Therefore, to get high input resistance, the bias resistance is bootstrapped.

6. Find out AC-coupled voltage follower? Answer: b [Reason:] An AC coupled voltage follower consists of coupling capacitor at the input of non-inverting terminal.

7. Voltage follower circuit are used in
a) Active filter
b) All of the mentioned
c) Sample and hold circuit
d) Bridge circuit with transducer

Answer: b [Reason:] Voltage followers are useful for all the above mentioned applications, because they involve working with high-input source resistance.

8. Compute the reference voltage for a fundamental log-amp, if it’s internal resistance=5MΩ.
a) 0.5µv
b) 0.05µv
c) 5µv
d) None of the mentioned

Answer: a [Reason:] Reference voltage, Vref = R1 × Is Where, Is~10-13A (for an emitter saturation current). ∴ Vref = 10-13 × 5MΩ = 5×10-7 = 0.5µv.

## Linear Integrated MCQ Set 2

1. Which device is used for diagnostic purposes and for recording?
a) Low pass filter
b) Monolithic PLL
c) Voltage Controlled Oscillator
d) None of the mentioned

Answer: c [Reason:] A Voltage Controlled Oscillator (VCO) is used for converting low frequency signals such as EEGs, EKG into an audio frequency range. These audio signals can be transmitted over two way radio communication systems for diagnostic purposes or can be recorded on a magnetic tape for further reference.

2. If the output of the Schmitt trigger is given below. Estimate the output at the pin 3 of VCO.  Answer: a [Reason:] In VCO, the output of Schmitt trigger is fed to the input of inverter. Therefore, the output at pin3 would be an inverted output. As the input is a square wave, the output obtained will be an inverted square wave.

3. Write the equation for time period of VCO?
a) (2×Vcc×CT)/i
b) (Vcc CT)/(2×i)
c) (Vcc×CT×i)/2
d) (2×Vcc)/(i×CT)

Answer: b [Reason:] The time period of VCO is given as T=2×△t =(2×0.25×Vcc ×CT)/i =(0.5 V×cc×CT)/i = (Vcc×CT)/(2×i).

4. Determine the value of current flow in VCO, when the NE566 VCO external timing resistor RT =250Ω and the modulating input voltage Vc=3.25V.(Assume Vcc=+5v).
a) 3mA
b) 12mA
c) 7mA
d) 10mA

Answer: c [Reason:] Current flowing in VCO, i =(Vcc– Vc)/ RT = (5V-3.25V)/250 = 1.75/250 =>i =7mA.

5. From the circuit given, find the value of output frequency? a) 178.484 Hz
b) 104.84 Hz
c) 145.84 Hz
d) 110.88 Hz

Answer: b [Reason:] Output frequency, fo =[2×(Vcc– Vc) ]/(CT×RT×Vcc )= [2x(8-1.5)]/(0.47µFx33kΩx8v) =13/0.124 => fo=104.84 Hz.

6. The output frequency of the VCO can be changed by changing
a) External tuning resistor
b) External tuning capacitor
c) Modulating input voltage
d) All of the mentioned

Answer: d [Reason:] The output frequency of VCO, fo = [2×(Vcc– Vc)]/(CT×RT×Vcc). From the equation, it is clear that the fo is inversely proportional to CT & RT and directly proportional to Vc.Therefore, the output frequency can be changed by changing either voltage control, CT or RT.

7. Calculate the value of external timing capacitor, if no modulating input signal is applied to VCO. Consider fo=25 kHz and RT=5 kΩ.
a) 6nF
b) 100µF
c) 2nF
d) 10nF

Answer: c [Reason:] When modulating input signal is not applied to VCO, the output frequency becomes fo=1/(4×RT×CT) => CT =1/(4×RT×fo) =1/(4×5kΩ×25kHz) = 2×10-9 =2nF.

8. What is the advantage of using filter?
a) High noise immunity
b) Reduce the bandwidth of PLL
c) Provides dynamic range of frequencies
d) None of the mentioned

Answer: a [Reason:] The charge on the filter capacitor gives a short time memory to the PLL. So, even if the signal becomes less than the noise for a few cycles, the dc voltage on the capacitor continues to shift the frequency of VCO, till it picks up the signal again. This produces high noise immunity.

9. Which filter is used in VCO? Answer: d [Reason:] The loop filter used in the VCO can be one of the three types of filter shown above.

10. Choose the VCO for attaining higher output frequency.
a) NE566
b) SE566
c) MC4024
d) All of the mentioned

Answer: c [Reason:] MC4024 is used for attaining high output frequency, because the maximum output frequency of NE566 and SE566 is 500kHz.

11. Voltage to frequency conversion factor for VCO is
a) Kv = △Vc/ △fo
b) Kv = △fo/△Vc
c) Kv = △fo × △Vc
d) Kv = 1/(△fo×△Vc)

Answer: b [Reason:] The voltage to frequency conversion factor is defined as the change in frequency to the change in modulating input voltage. => Kv=△fo/△Vc.

12. Calculate the voltage to frequency conversion factor, where fo=155Hz and Vcc=10V.
a) 130
b) 124
c) 134
d) 116

Answer: b [Reason:] The voltage to frequency conversion factor, Kv = △fo/△Vcc= 8×fo/Vcc = (8×155)/10=124.

13. Find the equation for change in frequency of VCO?
a) △fo = (2×△Vc)/(RT×CT×Vcc)
b) △fo = △Vc/(4×RT×CT×Vcc)
c) △fo = △Vc/(2×RT×CT×Vcc)
d) △fo = (4×△Vc)/(RT×CT×Vcc)

Answer: a [Reason:] The original output frequency, fo =2×[Vcc-Vc+△Vc]/[RT×CT×Vcc]. The new frequency f1 =2×[Vcc-Vc]/[RT×CT×Vcc]. Change in frequency, △fo= fo– f1 = 2×[Vcc-Vc+△Vc]/[RT×CT×Vcc]-{2×[Vcc-Vc]/[ RT×CT×Vcc] => △fo = (2×△Vc)/(RT×CT×Vcc).

14. Using the given specifications, determine the voltage to frequency conversion factor. a) 8.32
b) 8.90
c) 8.51
d) 8.75

Answer: c [Reason:] △fo = 2×△Vc/(RT×CT×Vcc) =>△Vc= (△fo×RT×CT×Vcc)/2 = (4.7µFx10kΩx5x112)/2 = 13.16V. Kv= △fo/ △Vc = 112Hz/13.16V =>Kv=8.51.

## Linear Integrated MCQ Set 3

1. Voltage to current converter is also called as
a) Current series positive feedback amplifier
b) Voltage series negative feedback amplifier
c) Current series negative feedback amplifier
d) Voltage series positive feedback amplifier

Answer: c [Reason:] Voltage to current converter is also called as current series negative feedback amplifier because the feedback voltage across internal resistor applied to the inverting terminal depends on the output current and is in series with the input difference voltage.

2. Given voltage to current converter with floating load. Determine the output current? a) 3mA
b) 6mA
c) 4mA
d) 2mA

Answer: d [Reason:] Output current, Io = Vin /R1 = 10/5kΩ =2mA.

3. Which of the following application uses voltage to current converter?
a) Low voltage dc and ac voltmeter
b) Diode match finding
c) Light emitting diode
d) All of the mentioned

Answer: d [Reason:] In all the applications mentioned above, the input voltage Vin is converted into an output current of Vin/R1 or the input voltage appear across resistor.

4. The op-amp in low voltage DC voltmeter cannot be nullified due to
a) D’Arsonaval meter movement
b) Offset voltage compensating network
c) Selection of switch
d) Gain of amplifier

Answer: a [Reason:] The op-amp sometimes cannot be nullified because the output is very sensitive to even slight variation in wiper position of D’Arsonaval meter movement (ammeter with a full scale deflection of 1mA).

5. What is the maximum input voltage that has to be selected to calibrate a dc voltmeter with a full scale voltage range of 1-13v.
a) ≤ ±14v
b) ≥ ±13v
c) ≤ ±15v
d) = ±14v

Answer: a [Reason:] The maximum input voltage has to be ≤ ±14v, to obtain the maximum full scale input voltage of 13v.

6. Higher input voltage can be measured in low voltage DC voltmeter using
a) Smaller resistance value
b) Higher resistance value
c) Random resistance value
d) All of the mentioned

Answer: b [Reason:] Higher resistance values are required to measure relatively higher input voltage. For example, if the range of switch is at x10 position in the low voltage dc voltmeter then, the corresponding resistance value would be 10kΩ. So, it requires a 10v input to get a full scale deflection (if 1v cause full scale deflection in the ammeter with a full scale deflection of 1mA).

7. In the diagram given below, determine the deflection of the ammeter with a full scale deflection of 1mA when the switch is at X2kΩ. Consider resistance of the offset voltage compensating network to be 10Ω. a) Full scale deflection in the ammeter
b) Half scale deflection in the ammeter
c) Quarter scale deflection in the ammeter
d) No deflection occurs in the ammeter

Answer: b [Reason:] Given Vin=1v ,R1=10+2kΩ ≅2kΩ Io = Vin/R1= 1v/2kΩ =0.5mA. This means that 2v causes half scale deflection of the ammeter.

8. How to modify a low voltage DC voltmeter to low voltage ac voltmeter
a) Add a full wave rectifier in the feedback loop
b) Add a half wave rectifier in the feedback loop
b) Add a square wave rectifier in the feedback loop
b) Add a sine wave rectifier in the feedback loop

Answer: a [Reason:] A combination of an ammeter and a full wave rectifier can be employed in the feedback loop to form an ac voltmeter.

## Linear Integrated MCQ Set 4

1. Which circuit converts irregularly shaped waveform to regular shaped waveforms?
a) Schmitt trigger
b) Voltage limiter
c) Comparator
d) None of the mentioned

Answer: a [Reason:] Schmitt trigger are also called as squaring circuit because, this type of circuit converts an irregularly shaped wave to a square wave or pulse.

2. Determine the upper and lower threshold voltage a) VUT = +14.63v, VLT= +14.63v
b) VUT = -14.63v, VLT= -14.63v
c) VUT = VLT= ±14.63v
d) None of the mentioned

Answer: b [Reason:] Upper threshold voltage, VUT = [R1/(R1+ R2)]× (+Vsat) = [10kΩ/(10kΩ +250Ω)]×(+15v)= +14.63v. Lower threshold voltage VLT = [R1/(R1+ R2)]×( -Vsat) = [10kΩ /(10kΩ+250Ω)]×(-15v)= -14.63v.

3. What happens if the threshold voltages are made longer than the noise voltages in schmitt trigger?
a) All the mentioned
b) Enhance the output signal
c) Reduce the transition effect
d) Eliminate false output transition

Answer: d [Reason:] In schmitt trigger, if the threshold voltage VUT and VLT are made larger than the input noise voltage. The positive feedback will eliminate the false output transition.

4. To a schmitt trigger in non-inverting configuration an input triangular wave of 1Vp is applied. What will be the output waveform, if the upper and lower threshold voltages are 0.25v?
a) Square waveform
b) Pulse waveform
c) Sawtooth waveform
d) Cannot be determined

Answer: a [Reason:] The input waveform has a threshold level of ±0.25v. 5. In which configuration a dead band condition occurs in schmitt trigger
a) Differential amplifier with positive feedback
b) Voltage follower with positive feedback
c) Comparator with positive feedback
d) None of the mentioned

Answer: c [Reason:] The comparator with positive feedback is said to exhibit hysteresis, a dead band condition, when the input of comparator exceeds upper threshold voltage. At this condition, output switch from +Vsat to -Vsat. It reverts back to its original state, +Vsat when the input goes below lower threshold voltage.

6. Calculate the hysteresis voltage for the schmitt trigger from the given specification:
R2 =56kΩ , R1 = 100Ω ,Vref = 0v & Vsat = ±14v. a) 0 mv
b) 25 mv
c) 50 mv
d) -25 mv

Answer: c [Reason:] Upper threshold voltage, VUT =[R1/(R1+R2)]×( +Vsat) = [100kΩ/(56kΩ +100 Ω)]×(+14v)= +25mv. Lower threshold voltage VLT = [R1/(R1+ R2)]×(-Vsat) = [100kΩ /(56kΩ+100Ω)]×(-14v)= -25 mv. ∴ Hysteresis voltage = VUT-VLT = 25-(-25) = 50mv.

7. How to limit the output voltage swing only to positive direction?
a) Combination of two zener diodes
b) Combination of zener and rectifier diode
c) All of the mentioned
d) Combination of two rectifier diodes

Answer: b [Reason:] To limit the output voltage swing to positive or negative direction, the basic op-amp comparator should be connected with a combination of zener and rectifier diode in the feedback path.

8. For the circuit shown below, obtain output waveform. Assume zener voltage to be 4.78v and voltage drop across the forward biased zener to be 0.7v.  Answer: a [Reason:] During positive half cycle of the input waveform, the output voltage is equal to (VZ +VD1) (because diode D1 would be in forward bias) , ∴ (VZ +VD1) = 4.78 v+0.7 v = 5.5 v. Similarly, during negative half cycle of the input wave form, the output voltage is equal to – (VZ +VD2) as the diode D2 would be in forward bias) , – (VZ +VD2) =- 4.78 v-0.7 v = -5.5 v. 9. A basic op-amp circuit has a zener and rectifier diode connected in the feedback path. Calculate the maximum positive voltage. Where, zener voltage = 5.1 v and voltage drop across the forward biased zener = 0.7v?
a) VO = 5.8v
b) VO = 9.9v
c) VO = 4.7v
d) VO = 7.1v

Answer: d [Reason:] Initially, rectifier diode will be reverse biased and makes the op-amp to operate in open loop configuration. So, the output voltage is obtained till the rectifier diode is forward bias and zener goes into avalanche condition. Hence, the maximum positive output voltage VOz +VD (VD –> voltage drop across rectifier diode). => VO= 5.1v+0.7 v= 5.8v.

10. Use the specification and obtain the output voltage swing for op-amp comparator.
Specification: R= 1kΩ; RL=10kΩ; VZ=6v; VSat=±15v (Assume forward bias of zener = 0.7v).  Answer: b [Reason:] During the positive half cycle, the output voltage would be at -VD = -0.7v because the zener will be forward biased. However, during negative half cycle of VO would be at +VZ =+6v. Thus, the zener diode in the feedback path limits VO to swing between +6v to -0.7v. ## Linear Integrated MCQ Set 5

1. Specify the voltage gain of non-inverting amplifier with feedback amplifier with and without feedback?
a) A= Vo/Vid, AF = Vf/Vo
b) A= Vf/Vid, AF= Vo/Vf
c) A= Vo/Vid, AF= Voin
d) A= Vf/Vid, AF = Vf/Vin

Answer: c [Reason:] The voltage gain of op-amp with feedback is the open loop voltage gain, A=Vo /Vid. The voltage gain of op-amp without feedback is the closed loop voltage gain, AF = Vo /Vin.

2. If the feedback voltage and the output voltage are given as 10v and 4v. Find the gain of the feedback circuit in voltage-series feedback amplifier?
a) 2.5v
b) 40v
c) 3v
d) 6.2v

Answer: a [Reason:] Gain of feedback, B = Vf /Vo = 10v/4v = 2.5v.

3. How is the difference voltage calculated in closed loop non-inverting amplifier?
a) Vid= Vo – Vf
b) Vid= Vin – Vf
c) Vid= Vo – Vin
d) Vid= Vf – Vin

Answer: b [Reason:] Although, the input is given to the non-inverting terminal of op-amp, the difference voltage is equal to the input voltage minus feedback voltage is Vid= Vin – Vf.

4. Why the feedback circuit is said to be negative for voltage series feedback amplifier?
a) Feedback voltage is 180o out of phase with respect to input voltage
b) Input voltage is 180o out of phase with respect to feedback voltage
c) Feedback voltage is in same phase with respect to input voltage
d) Input voltage is in same phase with respect to feedback voltage

Answer: a [Reason:] Voltage series feedback amplifier have the difference voltage, Vid = Vin-Vf. Therefore, the feedback voltage always opposes the input voltage and is out of phase by 180o with respect to input voltage. Hence, the feedback is said to be negative.

5. Determine the closed loop voltage gain from the given circuit. (Where gain of op-amp= 105). a) 1090.9
b) 9821.43
c) 9166.66
d) 10000

Answer: b [Reason:] The closed loop voltage gain , AF = {[A*(R4+R5)]/[(RF+R1+(A*R1)]} = [105*(10kΩ+1kΩ)]/[1kΩ+10kΩ+(105*1kΩ)] =11×108/112000 => AF = 9821.43.

6. Express closed loop voltage gain (AF) in terms of open loop gain (A) and feedback circuit gain (B)?
a) AF = A/AB
b) AF = 1+ (A/AB)
c) AF = A/(1+AB)
d) AF = AB/(1+A)

Answer: c [Reason:] The closed loop voltage gain in terms of open loop gain and feedback circuit gain is expressed as AF = A/(1+AB).

7. Which factor determines the gain of the voltage series feedback amplifier?
a) Open loop voltage gain
b) Feedback voltage
c) Ratio of two resistors
d) Gain of feedback circuit

Answer: c [Reason:] In setting the gain of the voltage series feedback amplifier, the ratio of two resistors is important and not the absolute value of these resistors. For example: If a gain of 11 is desired, we choose R1=1kΩ and R1=10kΩ or R1=100Ω and RF= 1kΩ.

8. For the feedback circuit of voltage series feedback amplifier, find the feedback voltage for the specifications: R1=1kΩ, RF = 10kΩ and Vo= 25v.
a) 12.5v
b) 22v
c) 0.9v
d) 2.3v

Answer: d [Reason:] The feedback voltage, Vf = (R1*Vo)/(R1+RF) = (1kΩ*25v)/(1kΩ+10kΩ) = 2.272v ≅ 2.3v.

9. What must be the value of external components used in voltage series feedback amplifier?
a) Less than 1MΩ
b) Less than 10MΩ
c) Less than 100MΩ
d) Less than 0.1MΩ

Answer: a [Reason:] All external components value should be less than 1MΩ. So, that they do not adversely affect the internal circuitry of the op-amp.

10. Find the block diagram representation of non-inverting amplifier with feedback? Answer: a [Reason:] The mentioned block diagram is the standard form for representing a system with feedback.

11. Define the input resistance with feedback for voltage series feedback amplifier?
a) RIF = (1-AB)
b) RIF = (AB-1)
c) RIF = (1+AB)
d) None of the mentioned